Question

If $2\tan 3A\cos 3A-\tan 3A+1=2\cos 3A$ and one value of A is ${20}^{\circ }$, the other value is:

Answer 1

$2\tan 3A\cos 3A-\tan 3A+1=2\cos 3A$
$2\tan 3A\cos 3A-\tan 3A-2\cos 3A+1=0$
$\tan 3A(2\cos 3A-1)-1(2\cos 3A-1)=0$
$(\tan 3A-1)(2\cos 3A-1)=0$
So,

$\tan 3A-1=0$ or $2\cos 3A-1=0$

$\tan 3A=1$ or $\cos 3A=\frac{1}{2}$

$\tan 3A=\tan {45}^{\circ }$ or $\cos 3A=\cos {60}^{\circ }$

$3A={45}^{\circ }$ or $3A={60}^{\circ }$

$A={15}^{\circ }$ or $A={20}^{\circ }$

Thus, $A={15}^{\circ }$ or ${20}^{\circ }.$